WBJEE · Maths · Functions
Let \(f: R \rightarrow R\) be defined by \(f(x)=x^{2}-\frac{x^{2}}{1+x^{2}}\) for all \(x \in R\). Then,
- A \(f\) is one-one but not onto mapping
- B \(f\) is onto but not one-one mapping
- C \(f\) is both one-one and onto
- D \(f\) is neither one-one nor onto
Answer & Solution
Correct Answer
(D) \(f\) is neither one-one nor onto
Step-by-step Solution
Detailed explanation
We have, \[ \begin{aligned} f(x) &=x^{2}-\frac{x^{2}}{1+x^{2}} \\ f(-x) &=(-x)^{2}-\frac{(-x)^{2}}{1+(-x)^{2}} \\ &=x^{2}-\frac{x^{2}}{1+x^{2}}=f(x) \\ \therefore f(-x)=f(x) & \end{aligned} \] So, \(f(x)\) is many one Again, \(f(x)=x^{2}-\frac{x^{2}}{1+x^{2}}\)…
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